Chapter 20 Rigid Body: Translation and Rotational Motion Kinematics
for Fixed Axis Rotation
20.1 Introduction ........................................................................................................... 1 20.2 Constrained Motion: Translation and Rotation ................................................ 1 20.2.1 Rolling without slipping ................................................................................ 5 Example 20.1 Bicycle Wheel Rolling Without Slipping ........................................ 6 Example 20.2 Cylinder Rolling Without Slipping Down an Inclined Plane ........ 9 Example 20.3 Falling Yo-Yo .................................................................................. 10 Example 20.4 Unwinding Drum ............................................................................ 11 20.3 Angular Momentum for a System of Particles Undergoing Translational and
Rotational .................................................................................................................... 12 Example 20.5 Earth’s Motion Around the Sun .................................................... 15 20.4 Kinetic Energy of a System of Particles ............................................................ 17 20.5 Rotational Kinetic Energy for a Rigid Body Undergoing Fixed Axis Rotation
...................................................................................................................................... 18 Appendix 20A Chasles’s Theorem: Rotation and Translation of a Rigid Body ... 19 Chapter 20 Rigid Body: Translation and Rotational Motion
Kinematics for Fixed Axis Rotation
Hence I feel no shame in asserting that this whole region engirdled by the
moon, and the center of the earth, traverse this grand circle amid the rest
of the planets in an annual revolution around the sun. Near the sun is the
center of the universe. Moreover, since the sun remains stationary,
whatever appears as a motion of the sun is really due rather to the motion
of the earth.1
Copernicus
20.1 Introduction
The general motion of a rigid body of mass m consists of a translation of the center of
mass with velocity Vcm and a rotation about the center of mass with all elements of the

rigid body rotating with the same angular velocity ω cm . We prove this result in Appendix
A. Figure 20.1 shows the center of mass of a thrown rigid rod follows a parabolic
trajectory while the rod rotates about the center of mass.
Figure 20.1 The center of mass of a thrown rigid rod follows a parabolic trajectory while
the rod rotates about the center of mass.
20.2 Constrained Motion: Translation and Rotation
We shall encounter many examples of a rolling object whose motion is constrained. For
example we will study the motion of an object rolling along a level or inclined surface
and the motion of a yo-yo unwinding and winding along a string. We will examine the
constraint conditions between the translational quantities that describe the motion of the
center of mass, displacement, velocity and acceleration, and the rotational quantities that
describe the motion about the center of mass, angular displacement, angular velocity and
angular acceleration. We begin with a discussion about the rotation and translation of a
rolling wheel.
1
Nicolaus Copernicus, De revolutionibus orbium coelestium (On the Revolutions of the Celestial Spheres),
Book 1 Chapter 10.
20-1
Figure 20.2 Rolling Wheel
Consider a wheel of radius R is rolling in a straight line (Figure 20.2).
 The center of
mass of the wheel is moving in a straight line at a constant velocity Vcm . Let’s analyze
the motion of a point P on the rim of the wheel.

Let v P denote the velocity of a point P on the rim of the wheel with respect to reference

frame O at rest with respect to the ground (Figure 20.3a). Let v ′P denote the velocity of
the point P on the rim with respect to the center of mass reference frame Ocm moving

with velocity Vcm with respect to at O (Figure 20.3b). (You should review the definition
of the center of mass reference frame in Chapter 15.2.1.) We can use the law of addition
of velocities (Eq.15.2.4) to relate these three velocities,



v P = v ′P + Vcm .
(20.2.1)
Let’s choose Cartesian coordinates for the translation motion and polar coordinates for
the motion about the center of mass as shown in Figure 20.3.
(a)
(b)
Figure 20.3 (a) reference frame fixed to ground, (b) center of mass reference frame
The center of mass velocity in the reference frame fixed to the ground is given by

Vcm = Vcm î .
(20.2.2)
20-2
where Vcm is the speed of the center of mass. The position of the center of mass in the
reference frame fixed to the ground is given by

R cm (t) = (Xcm, 0 + Vcm t) î ,
(20.2.3)
where Xcm, 0 is the initial x -component of the center of mass at t = 0 . The angular
velocity of the wheel in the center of mass reference frame is given by

ω cm = ω cm k̂ .
(20.2.4)
where ω cm is the angular speed. The point P on the rim is undergoing uniform circular
motion with the velocity in the center of mass reference frame given by

v ′P = Rω cm θ̂ .
(20.2.5)

If we want to use the law of addition of velocities then we should express v ′P = Rω cm θ̂ in
Cartesian coordinates. Assume that at t = 0 , θ (t = 0) = 0 i.e. the point P is at the top of
the wheel at t = 0 . Then the unit vectors in polar coordinates satisfy (Figure 20.4)
r̂ = sin θ î − cosθ ĵ
θ̂ = cosθ î + sin θ ĵ
.
(20.2.6)
Therefore the velocity of the point P on the rim in the center of mass reference frame is
given by

(20.2.7)
v ′P = Rω cm θ̂ = Rω cm (cosθ î − sin θ ĵ) .
Figure 20.4 Unit vectors
Now substitute Eqs. (20.2.2) and (20.2.7) into Eq. (20.2.1) for the velocity of a point P
on the rim in the reference frame fixed to the ground

v P = Rω cm (cosθ î + sin θ ĵ) + Vcm î
= (Vcm + Rω cm cosθ ) î + Rω cm sin θ ĵ
.
(20.2.8)
20-3
The point P is in contact with the ground when θ = π . At that instant the velocity of a
point P on the rim in the reference frame fixed to the ground is

v P (θ = π ) = (Vcm − Rω cm ) î .
(20.2.9)
What velocity does the observer at rest on the ground measure for the point on the rim
when that point is in contact with the ground? In order to understand the relationship
between Vcm and ω cm , we consider the displacement of the center of mass for a small
time interval Δt (Figure 20.5).
Figure 20.5 Displacement of center of mass in ground reference frame.
From Eq. (20.2.3) the x -component of the displacement of the center of mass is
ΔXcm =Vcm Δt .
(20.2.10)
The point P on the rim in the center of mass reference frame is undergoing circular
motion (Figure 20.6).
Figure 20.6: Small displacement of point on rim in center of mass reference frame.
20-4
In the center of mass reference frame, the magnitude of the tangential displacement is
given by the arc length subtended by the angular displacement Δθ = ω cm Δt ,
Δs = RΔθ = Rω cm Δt .
(20.2.11)
Case 1: if the x -component of the displacement of the center of mass is equal to the arc
length subtended by Δθ , then the wheel is rolling without slipping or skidding, rolling
without slipping for short, along the surface with
ΔXcm = Δs .
(20.2.12)
Substitute Eq. (20.2.10) and Eq. (20.2.11) into Eq. (20.2.12) and divide through by Δt .
Then the rolling without slipping condition becomes
Vcm = Rω cm ,
(rolling without slipping) .
(20.2.13)
Case 2: if the x -component of the displacement of the center of mass is greater than the
arc length subtended by Δθ , then the wheel is skidding along the surface with
ΔXcm > Δs .
(20.2.14)
Substitute Eqs. (20.2.10) and (20.2.11) into Eq. (20.2.14) and divide through by Δt , then
Vcm > Rω cm ,
(skidding) .
(20.2.15)
Case 3: if the x -component of the displacement of the center of mass is less than the arc
length subtended by Δθ , then the wheel is slipping along the surface with
ΔXcm < Δs .
(20.2.16)
Arguing as above the slipping condition becomes
Vcm < Rω cm ,
(slipping) .
(20.2.17)
20.2.1 Rolling without slipping
When a wheel is rolling without slipping, the velocity of a point P on the rim is zero
when it is in contact with the ground. In Eq. (20.2.9) set θ = π ,


v P (θ = π ) = (Vcm − Rω cm ) î = (Rω cm − Rω cm ) î = 0 .
(20.2.18)
This makes sense because the velocity of the point P on the rim in the center of mass
reference frame when it is in contact with the ground points in the opposite direction as
the translational motion of the center of mass of the wheel. The two velocities have the
20-5
same magnitude so the vector sum is zero. The observer at rest on the ground sees the
contact point on the rim at rest relative to the ground.
Thus any frictional force acting between the tire and the ground on the wheel is static
friction because the two surfaces are instantaneously at rest with respect to each other.
Recall that the direction of the static frictional force depends on the other forces acting on
the wheel.
Example 20.1 Bicycle Wheel Rolling Without Slipping
Consider a bicycle wheel of radius R that is rolling in a straight line without slipping.
The velocity
 of the center of mass in a reference frame fixed to the ground is given by
velocity Vcm . A bead is fixed to a spoke a distance b from the center of the wheel
(Figure 20.7). (a) Find the position, velocity, and acceleration of the bead as a function of
time in the center of mass reference frame. (b) Find the position, velocity, and
acceleration of the bead as a function of time as seen in a reference frame fixed to the
ground.
Figure 20.7 Example 20.1
Figure 20.8 Coordinate system for bead
in center of mass reference frame
Solution: a) Choose the center of mass reference frame with an origin at the center of the
wheel, and moving with the wheel. Choose polar coordinates (Figure 20.8). The z component of the angular velocity ω cm = dθ / dt > 0 . Then the bead is moving uniformly
in a circle of radius r = b with the position, velocity, and acceleration given by

rb′ = b r̂,


2
v ′b = bω cm θˆ, a ′b = −bω cm
r̂ .
(20.2.19)
Because the wheel is rolling without slipping, the velocity of a point on the rim of the
wheel has speed v′P = Rω cm . This is equal to the speed of the center of mass of the wheel
Vcm , thus
Vcm = Rω cm .
(20.2.20)
Note that at t = 0 , the angle θ = θ 0 = 0 . So the angle grows in time as
20-6
θ (t) = ω cm t = (Vcm / R)t .
(20.2.21)
The velocity and acceleration of the bead with respect to the center of the wheel are then
bVcm2
bVcm ˆ 

v ′b =
θ , a ′b = − 2 r̂ .
R
R
(20.2.22)
b) Define a second reference frame fixed to the ground with choice of origin, Cartesian
coordinates and unit vectors as shown in Figure 20.9.
Figure 20.9 Coordinates of bead in reference frame fixed to ground
Then the position vector of the center of mass in the reference frame fixed to the ground
is given by

(20.2.23)
R cm (t) = X cm î + R ĵ = Vcmt î + R ĵ .
The relative velocity of the two frames is the derivative


dR cm dXcm
Vcm =
=
î = Vcm î .
dt
dt
(20.2.24)
Because the center of the wheel is moving at a uniform speed the relative acceleration of
the two frames is zero,


dVcm 
A cm =
= 0.
(20.2.25)
dt
Define the position, velocity, and acceleration in this frame (with respect to the ground)
by

rb (t) = xb (t) î + yb (t) ĵ,


v b (t) = vb,x (t) î + vb,y (t) ĵ, a(t) = ab,x (t) î + ab,y (t) ĵ .
(20.2.26)
Then the position vectors are related by
20-7



rb (t) = R cm (t) + rb′(t) .
(20.2.27)
In order to add these vectors we need to decompose the position vector in the center of
mass reference frame into Cartesian components,

rb′(t) = b r̂(t) = bsin θ (t) î + bcosθ (t) ĵ .
(20.2.28)
Then using the relation θ (t) = (Vcm / R)t , Eq. (20.2.28) becomes



rb (t) = R cm (t) + rb′(t) = (Vcm t î + R ĵ) + (bsin θ (t) î + bcosθ (t) ĵ)
= (Vcm t + bsin((Vcm / R)t) ) î + ( R + bcos((Vcm / R)t) ) ĵ
.
(20.2.29)
Thus the position components of the bead with respect to the reference frame fixed to the
ground are given by
xb (t) = Vcm t + bsin((Vcm / R)t)
(20.2.30)
yb (t) = R + bcos((Vcm / R)t) .
(20.2.31)
A plot of the y -component vs. the x -component of the position of the bead in the
reference frame fixed to the ground is shown in Figure 20.10 below using the values
Vcm = 5 m ⋅ s-1 , R = 0.25 m , and b = 0.125 m . This path is called a cycloid. We can
differentiate the position vector in the reference frame fixed to the ground to find the
velocity of the bead

drb
d
d

v b (t) =
(t) = (Vcm t + bsin((Vcm / R)t)) î + (R + bcos((Vcm / R)t) ) ĵ ,
dt
dt
dt

v b (t) = (Vcm + (b / R)V cos((Vcm / R)t)) î − ((b / R)Vcm sin((Vcm / R)t) ) ĵ .
(20.2.32)
(20.2.33)
Figure 20.10 Plot of the y -component vs. the x -component of the position of the bead
20-8
Alternatively, we can decompose the velocity of the bead in the center of mass reference
frame into Cartesian coordinates

v ′b (t) = (b / R)Vcm (cos((Vcm / R)t) î − sin((Vcm / R)t) ĵ) .
(20.2.34)
The law of addition of velocities is then



v b (t) = Vcm + v ′b (t) ,

v b (t) = Vcm î + (b / R)Vcm (cos((Vcm / R)t) î − sin((Vcm / R)t) ĵ) ,
(20.2.35)
(20.2.36)

v b (t) = (Vcm + (b / R)Vcm cos((Vcm / R)t)) î − (b / R)sin((Vcm / R)t) ĵ , (20.2.37)
in agreement with our previous result. The acceleration is the same in either frame so


a b (t) = a ′b = −(b / R 2 )Vcm2 r̂ = −(b / R 2 )Vcm2 (sin((Vcm / R)t) î + cos((Vcm / R)t) ĵ) .
(20.2.38)
Example 20.2 Cylinder Rolling Without Slipping Down an Inclined Plane
A uniform cylinder of outer radius R and mass M with moment of inertia about the
center of mass I cm = (1 / 2) M R 2 starts from rest and rolls without slipping down an
incline tilted at an angle β from the horizontal. The center of mass of the cylinder has
dropped a vertical distance h when it reaches the bottom of the incline. Let g denote the
gravitational constant. What is the relation between the component of the acceleration of
the center of mass in the direction down the inclined plane and the component of the
angular acceleration into the page of Figure 20.11?
Figure 20.11 Example 20.2
Solution: We begin by choosing a coordinate system for the translational and rotational
motion as shown in Figure 20.12.
20-9
Figure 20.12 Coordinate system for rolling cylinder
For a time interval Δt , the displacement of the center of mass is given by

ΔR cm (t) = ΔX cm î . The arc length due to the angular displacement of a point on the rim
during the time interval Δt is given by Δs = RΔθ . The rolling without slipping condition
is
ΔX cm = RΔθ .
If we divide both sides by Δt and take the limit as Δt → 0 then the rolling without
slipping condition show that the x -component of the center of mass velocity is equal to
the magnitude of the tangential component of the velocity of a point on the rim
Vcm = lim
ΔX cm
Δt→0
Δt
= lim R
Δt→0
Δθ
= Rω cm .
Δt
Similarly if we differentiate both sides of the above equation, we find a relation between
the x -component of the center of mass acceleration is equal to the magnitude of the
tangential component of the acceleration of a point on the rim
Acm =
dVcm
dt
=R
dω cm
dt
= Rα cm .
Example 20.3 Falling Yo-Yo
A Yo-Yo of mass m has an axle of radius b and a spool of radius R (Figure 20.13a). Its
moment of inertia about the center of mass can be taken to be I = (1 / 2)mR 2 (the
thickness of the string can be neglected). The Yo-Yo is released from rest. What is the
relation between the angular acceleration about the center of mass and the linear
acceleration of the center of mass?
Solution: Choose coordinates as shown in Figure 20.13b.
20-10
Figure 20.13a Example 20.3
Figure 20.13b Coordinate system for
Yo-Yo
Consider a point on the rim of the axle at a distance r = b from the center of mass. As the
yo-yo falls, the arc length Δs = bΔθ subtended by the rotation of this point is equal to
length of string that has unraveled, an amount Δl . In a time interval Δt , bΔθ = Δl .
Therefore bΔθ / Δt = Δl / Δt . Taking limits, noting that, Vcm, y = dl / dt , we have that
bω cm = Vcm, y . Differentiating a second time yields bα cm = Acm, y .
Example 20.4 Unwinding Drum
Drum A of mass m and radius R is suspended from a drum B also of mass m and
radius R , which is free to rotate about its axis. The suspension is in the form of a
massless metal tape wound around the outside of each drum, and free to unwind (Figure
20.14). Gravity acts with acceleration g downwards. Both drums are initially at rest.
Find the initial acceleration of drum A , assuming that it moves straight down.
Figure 20.14 Example 20.4
Solution: The key to solving this problem is to determine the relation between the three
kinematic quantities α A , α B , and a A , the angular accelerations of the two drums and the
20-11
linear acceleration of drum A . Choose the positive y -axis pointing downward with the
origin at the center of drum B . After a time interval Δt , the center of drum A has
undergone a displacement Δy . An amount of tape Δl A = RΔθ A has unraveled from drum
A , and an amount of tape ΔlB = RΔθ B has unraveled from drum B . Therefore the
displacement of the center of drum A is equal to the total amount of tape that has
unwound from the two drums, Δy = Δl A + ΔlB = RΔθ A + RΔθ B . Dividing through by Δt
and taking the limit as Δt → 0 yields
dθ
dθ
dy
=R A+R B.
dt
dt
dt
Differentiating a second time yields the desired relation between the angular
accelerations of the two drums and the linear acceleration of drum A ,
d 2θ A
d 2θ B
d2y
=R 2 +R 2
dt 2
dt
dt
a A, y = Rα A + Rα B .
20.3 Angular Momentum for a System of Particles Undergoing
Translational and Rotational
We shall now show that the angular momentum of a body about a point S can be
decomposed into two vector parts, the angular momentum of the center of mass (treated
as a point particle) about the point S , and the angular momentum of the rotational
motion about the center of mass.
Consider a system of N particles located at the points labeled i = 1,2,, N . The angular
momentum about the point S is the sum
N 
N



Ltotal
=
L
=
(
∑ S ,i ∑ rS , i × mi v i ) ,
S
i=1
(20.3.1)
i=1

where rS , i is the vector from the point S to the ith particle (Figure 20.15) satisfying



rS , i = rS , cm + rcm,i ,



v S ,i = Vcm + v cm,i ,
(20.3.2)
(20.3.3)


where v S , cm = Vcm . We can now substitute both Eqs. (20.3.2) and (20.3.3) into Eq.
(20.3.1) yielding
N





Ltotal
=
(
r
+
r
)
×
m
(
V
+ v cm,i ) .
∑
S
S , cm
cm,i
i
cm
(20.3.4)
i=1
20-12
Figure 20.15 Vector Triangle
When we expand the expression in Equation (20.3.4), we have four terms,
N
N





Ltotal
=
(
r
×
m
v
)
+
∑ S ,cm i cm,i ∑ (rS ,cm × mi Vcm )
S
i=1
i=1
N




+ ∑ (rcm,i × mi v cm,i ) + ∑ (rcm,i × mi Vcm ).
N
i=1
(20.3.5)
i=1

The vector rS , cm is a constant vector that depends only on the location of the center of
mass and not on the location of the ith particle. Therefore in the first term in the above

equation, rS , cm can be taken outside the summation. Similarly, in the second term the

velocity of the center of mass Vcm is the same for each term in the summation, and may
be taken outside the summation,
 total 
⎛ N  ⎞ 
⎛ N ⎞
L S = rS ,cm × ⎜ ∑ mi v cm,i ⎟ + rS ,cm × ⎜ ∑ mi ⎟ Vcm
⎝ i=1
⎠
⎝ i=1 ⎠
⎛ N  ⎞ 


+ ∑ (rcm,i × mi v cm,i ) + ⎜ ∑ mircm,i ⎟ × Vcm .
⎝ i=1
⎠
i=1
N
(20.3.6)
The first and third terms in Eq. (20.3.6) are both zero due to the fact that
N

∑mr
i cm,i
=0
i=1
N

∑ mi v cm,i = 0.
(20.3.7)
i=1
N
We first show that

∑mr
i cm,i
is zero. We begin by using Eq. (20.3.2),
i=1
20-13
N
N
 
)
=
(mi (ri − rS ,cm ))
∑
cm,i

∑ (m r
i
i=1
i=1
N
⎞
 ⎛ N


= ∑ miri − ⎜ ∑ (mi )⎟ rS ,cm = ∑ miri − mtotal rS ,cm .
⎝ i=1
⎠
i=1
i=1
N
(20.3.8)
Substitute the definition of the center of mass (Eq. 10.5.3) into Eq. (20.3.8) yielding
N
N

)
=
mi ri − mtotal
∑
cm,i

∑ (m r
i
i=1
N
The vanishing of

∑m v
i
cm,i
i=1
1
mtotal
N


∑mr = 0.
i i
(20.3.9)
i=1
= 0 follows directly from the definition of the center of mass
i=1
frame, that the momentum in the center of mass is zero. Equivalently the derivative of
Eq. (20.3.9) is zero. We could also simply calculate and find that

∑m v
i
i
cm,i


= ∑ mi ( v i − Vcm )
i


= ∑ mi v i − Vcm ∑ mi
i
i


= mtotal Vcm − Vcm mtotal

= 0.
(20.3.10)
We can now simplify Eq. (20.3.6) for the angular momentum about the point S using the
N


fact that, mT = ∑ mi , and psys = mT Vcm (in reference frame O ):
i=1


 sys N 

Ltotal
=
r
×
p
+ ∑ (rcm,i × mi v cm,i ) .
S
S , cm
(20.3.11)
i=1



Consider the first term in Equation (20.3.11), rS ,cm × psys ; the vector rS ,cm is the vector
from the point S to the center of mass. If we treat the system as a point-like particle of
mass mT located at the center of mass, then the momentum of this point-like particle is


psys = mT Vcm . Thus the first term is the angular momentum about the point S of this
“point-like particle”, which is called the orbital angular momentum about S ,



Lorbital
= rS ,cm × psys .
S
(20.3.12)
for the system of particles.
20-14
N
Consider the second term in Equation (20.3.11),

∑ (r
cm,i

× mi v cm,i ) ; the quantity inside
i=1
th
the summation is the angular momentum of the i particle with respect to the origin in
the center of mass reference frame Ocm (recall the origin in the center of mass reference
frame is the center of mass of the system),



L cm,i = rcm,i × mi v cm,i .
(20.3.13)
Hence the total angular momentum of the system with respect to the center of mass in the
center of mass reference frame is given by
N 
N



Lspin
=
L
=
(rcm,i × mi v cm,i ) .
∑
∑
cm
cm,i
i=1
(20.3.14)
i=1
a vector quantity we call the spin angular momentum. Thus we see that the total angular
momentum about the point S is the sum of these two terms,

 orbital spin
Ltotal
=
L
+ L cm .
S
S
(20.3.15)
This decomposition of angular momentum into a piece associated with the translational
motion of the center of mass and a second piece associated with the rotational motion
about the center of mass in the center of mass reference frame is the key conceptual
foundation for what follows.
Example 20.5 Earth’s Motion Around the Sun
The earth, of mass me = 5.97 ×1024 kg and (mean) radius Re = 6.38 ×106 m , moves in a
nearly circular orbit of radius rs,e = 1.50 × 1011 m around the sun with a period
Torbit = 365.25 days , and spins about its axis in a period Tspin = 23 hr 56 min , the axis
inclined to the normal to the plane of its orbit around the sun by 23.5° (in Figure 20.16,
the relative size of the earth and sun, and the radius and shape of the orbit are not
representative of the actual quantities).
Figure 20.16 Example 20.5
20-15
If we approximate the earth as a uniform sphere, then the moment of inertia of the earth
about its center of mass is
2
I cm = me Re2 .
(20.3.16)
5
If we choose the point S to be at the center of the sun, and assume the orbit is circular,
then the orbital angular momentum is



Lorbital
= rS ,cm × psys = rs,e r̂ × me vcm θ̂ = rs,e me vcm k̂ .
S
(20.3.17)
The velocity of the center of mass of the earth about the sun is related to the orbital
angular velocity by
vcm = rs,e ω orbit ,
(20.3.18)
where the orbital angular speed is
ω orbit =
2π
2π
=
Torbit (365.25 d)(8.640 × 104 s ⋅ d −1 )
−7
(20.3.19)
−1
= 1.991 × 10 rad ⋅ s .
The orbital angular momentum about S is then

Lorbital
= me rs,2e ω orbit k̂
S
= (5.97 × 1024 kg)(1.50 × 1011 m)2 (1.991 × 10−7 rad ⋅ s −1 ) k̂
(20.3.20)
= (2.68 × 1040 kg ⋅ m 2 ⋅ s −1 ) k̂.
The spin angular momentum is given by


2
Lspin
= I cm ω spin = me Re2 ω spin n̂ ,
cm
5
(20.3.21)
where n̂ is a unit normal pointing along the axis of rotation of the earth and
ωspin =
2π
2π
=
= 7.293 ×10−5 rad ⋅ s −1 .
4
Tspin 8.616 ×10 s
(20.3.22)
The spin angular momentum is then

2
Lspin
= (5.97 × 1024 kg)(6.38 × 106 m)2 (7.293 × 10−5 rad ⋅ s −1 ) n̂
cm
5
= (7.10 × 1033 kg ⋅ m 2 ⋅ s −1 ) n̂.
(20.3.23)
20-16
The ratio of the magnitudes of the orbital angular momentum about S to the spin angular
momentum is greater than a million,
2
me rs,e2 ω orbit
Lorbital
5 rs,e Tspin
S
=
=
= 3.77 ×106 ,
2
2
Lspin
(2
/
5)
m
R
ω
2
R
T
cm
e e
spin
e
orbit
(20.3.24)
as this ratio is proportional to the square of the ratio of the distance to the sun to the
radius of the earth. The angular momentum about S is then

2
Ltotal
= me rs,e2 ω orbit kˆ + me Re2ωspin nˆ .
S
5
(20.3.25)
The orbit and spin periods are known to far more precision than the average values used
for the earth’s orbit radius and mean radius. Two different values have been used for one
“day;” in converting the orbit period from days to seconds, the value for the solar day,
Tsolar = 86, 400s was used. In converting the earth’s spin angular frequency, the sidereal
day, Tsidereal = Tspin = 86,160s was used. The two periods, the solar day from noon to noon
and the sidereal day from the difference between the times that a fixed star is at the same
place in the sky, do differ in the third significant figure.
20.4 Kinetic Energy of a System of Particles

Consider a system of particles. The ith particle has mass mi and velocity v i with respect
to a reference frame O . The kinetic energy of the system of particles is given by
1
1
 
K = ∑ mi vi2 = ∑ mi v i ⋅ v i
2 i
i 2


1


= ∑ mi ( v cm,i + Vcm ) ⋅( v cm,i + Vcm ).
2 i
(20.4.1)



where Equation 15.2.6 has been used to express v i in terms of v cm,i and Vcm . Expanding
the last dot product in Equation (20.4.1),



1



mi ( v cm,i ⋅ v i, rel + Vcm ⋅ Vcm + 2 v cm,i ⋅ Vcm )
∑
2 i



1
1



= ∑ mi ( v cm,i ⋅ v i, rel ) + ∑ mi ( Vcm ⋅ Vcm ) + ∑ mi v cm,i ⋅ Vcm
2 i
2 i
i
⎛
1
1
 ⎞ 
2
= ∑ mi vcm,i
+ ∑ mi Vcm2 + ⎜ ∑ m v cm,i ⎟ ⋅ Vcm .
⎝
⎠
2
2
K=
i
i
(20.4.2)
i
20-17
The last term in the third equation in (20.4.2) vanishes as we showed in Eq. (20.3.7).
Then Equation (20.4.2) reduces to
1
1
2
K = ∑ mi vcm,i
+ ∑ mi Vcm2
2 i
i 2
1
1
2
= ∑ mi vcm,i
+ mtotal Vcm2 .
2
i 2
(20.4.3)
We interpret the first term as the kinetic energy of the center of mass motion in reference
frame O and the second term as the sum of the individual kinetic energies of the particles
of the system in the center of mass reference frame Ocm .
At this point, it’s important to note that no assumption was made regarding the mass
elements being constituents of a rigid body. Equation (20.4.3) is valid for a rigid body, a
gas, a firecracker (but K is certainly not the same before and after detonation), and the
sixteen pool balls after the break, or any collection of objects for which the center of
mass can be determined.
20.5 Rotational Kinetic Energy for a Rigid Body Undergoing Fixed Axis
Rotation

The rotational kinetic energy for the rigid body, using v cm, i = (rcm, i )⊥ ω cm θ̂ , simplifies to
K rot =
1
2
I cmω cm
.
2
(20.5.1)
Therefore the total kinetic energy of a translating and rotating rigid body is
K total = K trans + K rot =
1
1
2
mVcm2 + I cmω cm
.
2
2
(20.5.2)
20-18
Appendix 20A Chasles’s Theorem: Rotation and Translation of a Rigid
Body
We now return to our description of the translating and rotating rod that we first
considered when we began our discussion of rigid bodies. We shall now show that the
motion of any rigid body consists of a translation of the center of mass and rotation about
the center of mass.
We shall demonstrate this for a rigid body by dividing up the rigid body into point-like
constituents. Consider two point-like constituents with masses m1 and m2 . Choose a

coordinate system with a choice of origin such that body 1 has position r1 and body 2

has position r2 (Figure 20A.1). The relative position vector is given by

 
r1,2 = r1 − r2 .
(20.A.1)
Figure 20A.1 Two-body coordinate system.

Recall we defined the center of mass vector, R cm , of the two-body system as



m1 r1 + m2 r2
.
R cm =
m1 + m2
(20.A.2)
In Figure 20A.2 we show the center of mass coordinate system.
Figure 20A.2 Position coordinates with respect to center of mass
20-19
The position vector of the object 1 with respect to the center of mass is given by


m2
µ 

 
 m1 r1 + m2 r2
 
rcm,1 = r1 − R cm = r1 −
=
(r1 − r2 ) =
r ,
m1 + m2
m1 + m2
m1 1,2
(20.A.3)
where
µ=
m1m2
,
m1 + m2
(20.A.4)
is the reduced mass. In addition, the relative position vector between the two objects is
independent of the choice of reference frame,



 





r12 = r1 − r2 = (rcm,1 + R cm ) − (rcm,2 + R cm ) = rcm,1 − rcm,2 = rcm,1,2 .
(20.A.5)
Because the center of mass is at the origin in the center of mass reference frame,


m1rcm,1 + m2rcm,2
m1 + m2
Therefore

= 0.


m1rcm,1 = −m2rcm,2


m1 rcm,1 = m2 rcm,2 .
(20.A.6)
(20.A.7)
(20.A.8)
The displacement of object 1 about the center of mass is given by taking the derivative of
Eq. (20.A.3),
µ 

(20.A.9)
drcm,1 =
dr .
m1 1,2
A similar calculation for the position of object 2 with respect to the center of mass yields
for the position and displacement with respect to the center of mass
µ 

 
rcm,2 = r2 − R cm = −
r ,
m2 1,2
(20.A.10)
µ 

drcm,2 = −
dr .
m2 1,2
(20.A.11)
Let i = 1,2 . An arbitrary displacement of the ith object is given respectively by



dri = drcm,i + dR cm ,
(20.A.12)
20-20

which is the sum of a displacement about the center of mass drcm,i and a displacement of

the center of mass dR cm . The displacement of objects 1 and 2 are constrained by the
condition that the distance between the objects must remain constant since the body is
rigid. In particular, the distance between objects 1 and 2 is given by
 2
 
 
r1,2 = (r1 − r2 ) ⋅ (r1 − r2 ) .
(20.A.13)
Because this distance is constant we can differentiate Eq. (20.A.13), yielding the rigid
body condition that
 




0 = 2(r1 − r2 ) ⋅ (dr1 − dr2 ) = 2r1,2 ⋅ dr1,2
(20.A.14)
20A.1. Translation of the Center of Mass
The condition (Eq. (20.A.14)) can be satisfied if the relative displacement vector between
the two objects is zero,


 
dr1,2 = dr1 − dr2 = 0 .
(20.A.15)
This implies, using, Eq. (20.A.9) and Eq. (20.A.11), that the displacement with respect to
the center of mass is zero,



drcm,1 = drcm,2 = 0 .
(20.A.16)
Thus by Eq. (20.A.12), the displacement of each object is equal to the displacement of
the center of mass,


dri = dR cm ,
(20.A.17)
which means that the body is undergoing pure translation.
20A.2 Rotation about the Center of Mass


 
Now suppose that dr1,2 = dr1 − dr2 ≠ 0 . The rigid body condition can be expressed in
terms of the center of mass coordinates. Using Eq. (20.A.9), the rigid body condition (Eq.
(20.A.14)) becomes
µ 

(20.A.18)
0=2
r1,2 ⋅ drcm,1 .
m1
Because the relative position vector between the two objects is independent of the choice
of reference frame (Eq. (20.A.5)), the rigid body condition Eq. (20.A.14) in the center of
mass reference frame is then given by


0 = 2rcm,1,2 ⋅ drcm,1 .
(20.A.19)
20-21
This condition is satisfied if the relative displacement is perpendicular to the line passing
through the center of mass,


rcm,1,2 ⊥ drcm,1 .
(20.A.20)


By a similar argument, rcm,1,2 ⊥ drcm,2 . In order for these displacements to correspond to a
rotation about the center of mass, the displacements must have the same angular
displacement.
Figure 20A.3 Infinitesimal angular displacements in the center of mass reference frame
In Figure 20A.3, the infinitesimal angular displacement of each object is given by

drcm,1
dθ1 = 
,
rcm,1

drcm,2
dθ 2 = 
.
rcm,2
(20.A.21)
(20.A.22)
From Eq. (20.A.9) and Eq. (20.A.11), we can rewrite Eqs. (20.A.21) and (20.A.22) as

µ dr1,2
dθ1 =
,
m1 rcm,1

µ dr1,2
dθ 2 =
.
m2 rcm,2
(20.A.23)
(20.A.24)


Recall that in the center of mass reference frame m1 rcm,1 = m2 rcm,2 (Eq. (20.A.8))
and hence the angular displacements are equal,
20-22
dθ1 = dθ 2 = dθ .
(20.A.25)

Therefore the displacement of the ith object dri differs from the displacement of the

center of mass dR cm by a vector that corresponds to an infinitesimal rotation in the center
of mass reference frame



dri = dR cm + drcm,i .
(20.A.26)
We have shown that the displacement of a rigid body is the vector sum of the
displacement of the center of mass (translation of the center of mass) and an infinitesimal
rotation about the center of mass.
20-23